(2x-3)3=(2x-3)(2x-3)2

Solution for (2x-3)3=(2x-3)(2x-3)2 equation:

(2x-3)3=(2x-3)(2x-3)2

We move all terms to the left:

(2x-3)3-((2x-3)(2x-3)2)=0

We multiply parentheses

6x-((2x-3)(2x-3)2)-9=0

We multiply parentheses ..

-((+4x^2-6x-6x+9)2)+6x-9=0


We calculate terms in parentheses: -((+4x^2-6x-6x+9)2), so:

(+4x^2-6x-6x+9)2

We multiply parentheses

8x^2-12x-12x+18

We add all the numbers together, and all the variables

8x^2-24x+18

Back to the equation:

-(8x^2-24x+18)


We add all the numbers together, and all the variables

6x-(8x^2-24x+18)-9=0

We get rid of parentheses

-8x^2+6x+24x-18-9=0

We add all the numbers together, and all the variables

-8x^2+30x-27=0

a = -8; b = 30; c = -27;
Δ = b2-4ac
Δ = 302-4·(-8)·(-27)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-6}{2*-8}=\frac{-36}{-16}
=2+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+6}{2*-8}=\frac{-24}{-16}
=1+1/2
$